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2001 AIME II Problems/Problem 4

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Problem

Let R = (8,6). The lines whose equations are 8y = 15x and 10y = 3x contain points P and Q, respectively, such that R is the midpoint of \overline{PQ}. The length of PQ equals \frac {m}{n}, where m and n are relatively prime positive integers. Find m + n.

Solution

pointpen = black; pathpen = black+linewidth(0.7);pair R = (8,6), P = (32,60)/7, Q= (80,24)/7;D((0,0)--MP("x",(13,0)...

The coordinates of P can be written as \left(a, \frac{15a}8\right) and the coordinates of point Q can be written as \left(b,\frac{3b}{10}\right). By the midpoint formula, we have \frac{a+b}2=8 and \frac{15a}{16}+\frac{3b}{20}=6. Solving for b gives b= \frac{80}{7}, so the point Q is \left(\frac{80}7, \frac{24}7\right). The answer is twice the distance from Q to (8,6), which by the distance formula is \frac{60}{7}. Thus, the answer is \boxed{067}.

See also

2001 AIME II (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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